Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
\cos A = \frac{{A{B^2} + A{C^2} - B{C^2}}}{{2.AB.AC}} = \frac{{{7^2} + {8^2} - {{13}^2}}}{{2.7.8}} = - \frac{1}{2}\\
\Rightarrow \widehat A = 120^\circ
\end{array}\)
b,
Ta có:
\(\begin{array}{l}
\sin \widehat A = \sin 120^\circ = \frac{{\sqrt 3 }}{2}\\
{S_{ABC}} = \frac{1}{2}AB.AC.\sin A = \frac{1}{2}.7.8.\frac{{\sqrt 3 }}{2} = 14\sqrt 3
\end{array}\)
c,
\(BH = \frac{{2{S_{ABC}}}}{{AC}} = \frac{{2.14\sqrt 3 }}{8} = \frac{{7\sqrt 3 }}{2}\)